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DOMES OVER CURVES

ALEXEY GLAZYRIN⋆ AND IGOR PAK⋄

Abstract. A closed piecewise linear curve is called integral if it is comprised of unit intervals. Kenyon’sproblem asks whether for every integral curve γ in R3, there is a dome over γ, i.e. whether γ is a boundaryof a polyhedral surface whose faces are equilateral triangles with unit edge lengths. First, we give analgebraic necessary condition when γ is a quadrilateral, thus giving a negative solution to Kenyon’sproblem in full generality. We then prove that domes exist over a dense set of integral curves. Finally,we give an explicit construction of domes over all regular n-gons.

1. Introduction

The study of polyhedra with regular polygonal faces is a classical subject going back to ancient times.It was revived periodically when new tools and ideas have developed, most recently in connection toalgebraic tools in rigidity theory. In this paper we study one of most basic problems in the subject –polyhedral surfaces in R3 whose faces are congruent equilateral triangles. We prove both positive andnegative results on the types of boundaries these surfaces can have, suggesting a rich theory extendingfar beyond the current state of the art.

Formally, let γ ⊂ R3 be a closed piecewise linear (PL-) curve.1 We say that γ is integral if it iscomprised of intervals of integer length. Now, let S ⊂ R3 be a PL-surface realized in R3 with theboundary ∂S = γ, and with all facets comprised of unit equilateral triangles. In this case we say that Sis a unit triangulation or dome over γ, that γ is spanned by S, and that γ can be domed. By a PL-surfacewe mean a realization of a pure connected finite 2-dimensional simplicial complex, with no additionalrestriction of embedding or immersion.

Question 1.1 (Kenyon, see §6.2). Is every integral closed curve γ ⊂ R3 spanned by a unit triangulation?In other words, can every such γ be domed?

For example, the unit square and the (unit sided) regular pentagon can be domed by a regular pyramidwith triangular faces. Of course, there is no such simple construction for a regular heptagon. Perhapssurprisingly, the answer to Kenyon’s question is negative in general.

A 3-dimensional unit rhombus is a closed curve ρ ⊂ R3 with four edges of unit length. The unitrhombi form a 2-parameter family of space quadrilaterals ρ(a, b) parametrized by the diagonals a and b,defined as distances between pairs of opposite vertices.

Theorem 1.2. Let ρ(a, b) ⊂ R3 be a unit rhombus with diagonals a, b > 0. Suppose ρ(a, b) can bedomed. Then there is a nonzero polynomial P ∈ Q[x, y], such that P (a2, b2) = 0.

In other words, for a, b > 0 algebraically independent over Q, the corresponding unit rhombus cannotbe domed, giving a negative answer to Kenyon’s question. In fact, our tools give further examples ofunit rhombi which cannot be domed, such as ρ

(1π, 1π

), see Corollary 4.9.

The following result is a positive counterpart to the theorem. We show that the set of integral curvesspanned by a unit triangulation is everywhere dense within the set of all integral curves, with respect tothe topology induced by Frechet distance, see below.

⋆School of Mathematical & Statistical Sciences, University of Texas Rio Grande Valley, Brownsville, TX 78520. Email:

alexey.glazyrin@utrgv.edu.⋄Department of Mathematics, UCLA, Los Angeles, CA, 90095. Email: pak@math.ucla.edu.1To avoid unnecessary technical difficulties, we do not consider curves with different vertices mapped to the same point

although all our results remain true for them as well.

1

2 ALEXEY GLAZYRIN AND IGOR PAK

Let γ, γ′ ⊂ R3 be two integral closed curves of equal length. We assume the vertices of γ, γ′ aresimilarly labeled

[v1 . . . vn

]and

[v′1 . . . v

′n

], giving parametrizations of the curves. The Frechet distance

|γ, γ′|F in this case is given by

|γ, γ′|F = max1≤i≤n

|viv′i| ,

where |viv′i| is the Euclidean distance between vi and v′i.

Theorem 1.3. For every integral curve γ ⊂ R3 and ε > 0, there is an integral curve γ′ ⊂ R3 of equallength, such that |γ, γ′|F < ε and γ′ can be domed.

The theorem above does not give a concrete characterization of domed integral curves, and such acharacterization seems difficult (see §5). We conclude with one interesting special case:

Theorem 1.4. Every regular integral n-gon in the plane can be domed.

This gives a new infinite class of regular polygon surfaces, comprised of one regular n-gon and manyunit triangles. See Section 3 for the proof and some previously known special cases.

Outline of the paper. We begin with a technical proof of Theorem 1.3 in Section 2. Our proofis constructive and almost completely self-contained except for the Steinitz Lemma with Bergstromconstant, see §2.7. In Section 3, we follow with a (much shorter) constructive proof of Theorem 1.4,which is almost completely independent of the previous section, except for the earlier analysis of rhombiwhich can be domed, see §2.3.

In Section 4, we prove Theorem 1.2 by extending the results of Gaifullin brothers [13]. We assumethat the reader is familiar with the theory of places, see e.g. [21, Ch. 1] and [26, §41.7], which played akey role in the solution of the bellows conjecture, see [7] (see also [26, §34]). Shifting gears once again,Section 5 is independent of the rest of the paper. Here we make a number of interrelated conjectures onthe integral curves which can be domed, which we then relate to the rigidity theory and the EuclideanRamsey theory. Final remarks are given in Section 6.

In the Appendix A, we include a negative solution of the question in [13] on the dimension of theflexes of doubly periodic surfaces. This counterexample arose upon careful inspection of our proof ofTheorem 1.2, and is of independent interest (cf. [29]).

Notation. Let |vw| denote the Euclidean distance between v, w ∈ R3. We use [v1 . . . vn], vi ∈ Rd,to denote a closed polygonal curve γ ⊂ R3 with vertices v1, . . . , vn. We use parentheses notation(a1, . . . , an), ai > 0, to denote the edge lengths of γ, i.e. ai = |vivi+1|, and an = |vnv1|. Denote by|γ| = a1 + . . .+ an ∈ N the length of the integral curve γ.

Throughout the paper all curves will be integral and closed PL-curves in R3, unless stated otherwise.Similarly, all PL-surfaces S will have unit triangles, unless stated otherwise. They are realized in R3

by the vertex coordinates and such realizations have no additional extrinsic constraints (such as beingembedding or immersion, cf. §6.2).

2. Integral curves which can be domed are dense

2.1. Understanding domes over curves. Before proceeding to the proofs of the theorems, let us givesome basic ideas of domes over curves, and how they can be built. In Figure 1 two regular pentagonalpyramids give an example of domes over a regular pentagon. It’s really the same dome up to a rigidmotion. Similarly, the regular pentagonal biprism on the right can also be viewed as a dome overa disconnected integral curve comprised of two pentagons. We will not consider disconnected curvesuntil §5.5.

Now, arrange the pyramid and the biprism as in the figure and notice that they can be attached toeither of them to form yet another example of a dome over a pentagon. The idea of attaching triangulatedPL-surfaces will be used repeatedly through the paper, both for explicit constructions and to disproveexistence of other domes.

DOMES OVER CURVES 3

Figure 1. Three parts of an icosahedron as domes.

2.2. Mapping the proof of Theorem 1.3. The basic idea is to show that “generic” curves can besimplified to curves which can then be broken into pieces, each of which can have an explicit constructionof a dome. The process of simplifying the curve and the construction are sufficiently robust to allowreversing the process. In essence we first target a curve, and construct a “generic” curve that is closeenough.

Formally, denote by Mn the space of all integral curves of length n in R3, modulo rigid motions,which is compact in the Frechet topology. Let Dn ⊂Mn denote the subset of integral curves which canbe domed. The goal of this section is to prove Theorem 1.3, which states that Dn is dense inMn.

The proof goes through several stages of simplification of integral curves, along the following route:

integral curves −→ generic curves −→ near planar curves −→ compact near planar curves.

Compact curves are curves which fit inside a ball of radius 3/2 and they are much simpler to analyze byinduction. At each stage, the simplification of curves is made by a sequence of certain local transforma-tions. Starting the second arrow, these transformations are called flips and are obtained by attachingunit rhombi which can be domed. Making these reductions rigorous is somewhat technical and willoccupy much of this section. The rhombi ρ ∈ D4 will play a special role, so we consider them first.

2.3. Dense rhombi. Throughout the paper, a unit closed curve of length 4 is called a unit rhombus,or just a rhombus. Each unit rhombus is determined by the diagonals a and b; we denote such unitrhombus by ρ(a, b). Observe that a2 + b2 ≤ 4, with the equality achieved on plane rhombi.

Lemma 2.1. Fix the diagonal a, and suppose 0 < a < 2, a /∈ Q. Then the set of values of b ≥ 0 forwhich ρ(a, b) ∈ D4 is dense in

[0,√4− a2

].

Proof. Consider a planar isosceles trapezoid with side lengths (1, 1, 1, a). Its circumradius is 1√3−a

. Take

a line through the circumcenter orthogonal to the plane containing the trapezoid and choose two pointson this line with distance 1 from all vertices of the trapezoid (see Figure 2). Connecting these two pointsto the vertices, we obtain six unit triangles and a unit rhombus ρ1 spanned by them. The diagonals of

ρ1 are a and c1 = 2√

2−a3−a

.

ρ1

1

1

1

a

Figure 2. Construction of unit rhombus ρ1 in the proof of Lemma 2.1.

We glue two copies of ρ1, with a common diagonal of length a, via two of their sides. Define ρ2 =ρ(a, c2) to be a rhombus obtained as a boundary of the two glued copies of ρ1. Similarly, define ρ3 =ρ(a, c3) by gluing three copies of ρ1, etc. Clearly, every rhombus ρm, m ≥ 1, can be domed by surfacewith 6m unit triangles. We have:

cm =√4− a2 · | sinmα| , where α := arcsin

2√(2 + a)(3− a)

.

4 ALEXEY GLAZYRIN AND IGOR PAK

Thus, set cm,m ≥ 1 is dense in[0,√4− a2

], for all α /∈ πQ. Finally, we have α /∈ πQ, since otherwise

sinα = 2√(2+a)(3−a)

∈ Q, a contradiction with the assumption that a /∈ Q.

Let D′4 denote the set of unit rhombi that can be domed using the construction from Lemma 2.1. The

integer m in the proof will be called a multiplier throughout this section. We can now prove Theorem 1.3for |γ| = 4. Let

(2.1) X :=

x > 0 : arcsin

2√(2 + x)(3 − x)

∈ πQ

.

Lemma 2.2. Let ρ = ρ(a, b) ⊂ R3 be a unit rhombus, and let ε > 0. Then there is a unit rhombusρ′ = ρ(a′, b′) ∈ D4, such that |ρ, ρ′|F < ε. Moreover, if a /∈ X , one can take a′ = a.

Proof. The second part follows from the above proof of Lemma 2.1. For the first part, choose a /∈ X , sothat |a− a′| < ε. Then apply the construction as above.

2.4. Reachable curves. Let us introduce some definitions and notation. Consider two integral curvesγ = [v1 . . . vk . . . vn] and γ′ = [v1 . . . vk−1v

′kvk+1 . . . vn], such that [vk−1vkvk+1v

′k] ∈ D′

4. In this casewe say that γ and γ′ are k-flip connected, or just flip connected ; write γ →k γ

′. Two integral curvesγ = [v1 . . . vn] and γ′ = [v′1 . . . v

′n] are called flip equivalent, write γ ∼ γ′, if

(2.2) γ = γ0 →k1γ1 →k2

γ2 →k3. . . →kN

γN = γ′ ,

for some integer sequence k := (k1, . . . , kN ), where 1 ≤ ki ≤ N for all i = 1, . . . , N . See an example inFigure 3. Clearly, if γ ∼ γ′ and γ′ ∼ γ′′, then γ ∼ γ′′.

v1

γ γ γ0 1 2

v1 v1

v2

v2'

v3v3

v3'v2'

2 3

Figure 3. Sequence of two flips γ0 →2 γ1 →3 γ2.

We say that an integral curve γ ⊂ R3 of length n is reachable, if for all ε > 0, there is an integralcurve γ′ ⊂ R3 of length n, such that |γ, γ′|F < ε, and γ′ ∈ Dn. In this notation, Theorem 1.3 claimsthat all integral curves are reachable, while Lemma 2.2 proves this for curves of length 4.

Lemma 2.3. Let γ ∼ γ′ are flip equivalent integral curves in R3. Suppose γ is reachable. Then so is γ′.

In other words, the lemma says that if γ ∈ Mn is a limit point of Dn, then so are all flip equivalentcurves γ′ ∼ γ.Proof. Since γ ∼ γ′, there is a flip sequence k as in (2.2), and a sequence of multipliers m = (m1, . . . ,mN ) ∈Zm counting how many pairs of unit triangles are added at each flip. Use positive and negative inte-gers mi to denote clockwise or counterclockwise direction of the flip γi−1 → γi. A flip consists ofattaching sequentially m sets of six triangles. Assume the first rhombus in a sequence spanned by sixunit triangles is [vk−1vkvk+1v

′]. We say that the direction of the flip is positive if the determinant of thematrix (vk−1 − vk; v′ − vk; vk+1 − vk) is positive. Otherwise, we say the direction of the flip is negative.Thus, pair (k,m) uniquely encodes the combinatorial structure of the flip equivalence. Let

(2.3) Φk,m :Mn →Mn

be the map defining flip sequence as above. By construction, Φk,m : Dn → Dn, and Φk,m(γ) = γ′.Clearly, the map Φk,m is a composition of |m1|+. . .+|mN | continuous maps, and thus also continuous

on Mn. Since γ is reachable, there is a sequenceγ〈t〉 → γ, t ∈ N

of converging curves γ〈t〉 ∈ Dn.

Thus, we have another sequence of converging curves in Dn:Φk,m

(γ〈t〉

)→ γ′, t ∈ N

,

which shows that γ′ is reachable.

DOMES OVER CURVES 5

2.5. Generic curves. Let γ = [v1 . . . vn] be an integral curve in R3. Following [7, 27] (see also [26,§34]), the distances |vivi+2| are called small diagonals. Here and below, we have vertex indices 1 ≤ i ≤ n,and (i + 2) is taken modulo n.

We say that an integral curve γ ⊂ R3 is generic, if for all flip equivalent curves γ′ ∼ γ, whereγ′ = [v′1 . . . v

′n], we have small diagonals |v′iv′i+2| are not in Q. Denote by Gn ⊂ Mn the set of generic

integral curves of length n. By definition, if a curve is generic then its flip equivalent curve is also generic.

Lemma 2.4. Set Gn is dense in Mn.

Proof. A neighborhood of a given curve γ in Mn is a semi-algebraic set of dimension 2n. Assume aconcrete small diagonal |v′kv′k+2| of a curve γ′ is a fixed number from Q. The space of such curves in theneighborhood of γ is a semi-algebraic set of dimension at most 2n− 1. For a given sequence of flips, thespace of curves produced from γ′ with this sequence of flips is at most (2n−1)-dimensional as well. Sinceboth Q and the set of flip sequences are countable, the set of non-generic curves in the neighborhood ofγ is a union of countably many semi-algebraic sets of dimension at most 2n − 1. Therefore, there is acurve from Gn in an arbitrary neighborhood of γ.

2.6. Planar curves. An integral curve γ ∈ Mn is called planar if it lies in a plane H ⊂ R3. Denoteby Pn ⊂Mn the set of planar integral curves of length n.

Lemma 2.5. For the flip equivalence class of every γ ∈ Gn, there is a planar curve ξ ∈ Pn that is itslimit point.

In other words, for every ε > 0, and every generic integral curve γ ∈ Gn, there is a generic integralcurve γ′ ∈ Gn and a planar integral curve ξ ∈ Pn, such that γ ∼ γ′ and |γ′, ξ|F < ε. Note that thecurve ξ does not have to be generic itself, or be flip equivalent to γ.

Proof. The proof is based on the same idea of using flips to obtain a near-planar curve γ′. Let γ =[v1 . . . vn] ∈ Gn, and let ϕ : R3 → R be a generic linear function, i.e. such that for any curve flipequivalent to γ, values ϕ are different on all its vertices. Generic linear functions exist because there areonly countably many curves flip equivalent to a given one. Cyclically, for all k from 1 to n, make k-flips:

(2.4) γ = γ0 →1 γ1 →2 γ2 →3 . . . →n γn →1 γn+1 →2 γn+2 →3 . . .

Choose integers m (see the proof of Lemma 2.3), as follows. Consider a flip

γjn+k−1 = [. . . wk−1wkwk+1 . . .] →k γjn+k = [. . . wk−1w′kwk+1 . . .] .

By the proof of Lemma 2.1, we can always choose a multiplier mjn+k so that

(2.5)2

3α +

1

3β < ϕ(w′

k) <1

3α +

2

3β ,

where

α := minϕ(wk−1), ϕ(wk+1)

, β := max

ϕ(wk−1), ϕ(wk+1)

.

Note that we have α 6= β, since ϕ is generic, so there is always room to make such flip possible.Using (2.5), it is easy to see that there is a limit

(ϕ(w1), . . . , ϕ(wn)

)→ (h, . . . , h), for some h ∈ R,

Here the limit is when N →∞, where N is the number of flips in (2.4). Indeed, note that maxk ϕ(wk) isnon-increasing and thus converges to some M < ∞. Similarly, note that mink ϕ(wk) is non-decreasingand thus converges to some µ ≤ M . If µ = M , then we can take h = M = µ. Therefore, it remains toshow that µ < M is impossible.

Since maxk ϕ(wk) → M , there is a moment when all values of ϕ(wk) are smaller than M + δ for agiven δ. On the other hand there is ℓ such that ϕ(wℓ) 6= µ. Let us consider the next n flips startingfrom the vertex ℓ + 1. By the construction of the flip sequence, we have ϕ(w′

ℓ+1) <13 µ + 2

3 (M + δ),

then ϕ(w′ℓ+2) <

19 µ + 8

9 (M + δ), etc. We conclude that all values of ϕ after n flips are no greater than13n µ + 3n−1

3n (M + δ). For µ < M , we can choose δ > 0 such that this value is smaller than M . Thiscontradicts the assumption maxk ϕ(wk)→M , and implies µ =M .

The limit curve ξ is integral and lies in the plane H := x ∈ R3 : ϕ(x) = h. Therefore, forN = N(ε) large enough, we obtain a curve γ′ := γN , such that γ′ ∼ γ, and |γ′, ξ|F < ε.

6 ALEXEY GLAZYRIN AND IGOR PAK

2.7. Packing curves. Let u1, . . . , un ∈ Rd be unit vectors which satisfy u1+ . . .+un = 0. The SteinitzLemma famously states, see e.g. [3], that there is always a permutation σ ∈ Sn, s.t.

(2.6)∣∣uσ(1) + . . . + uσ(k)

∣∣ ≤ Bd for all 1 ≤ k ≤ n,where Bd ≤ 2d is a universal constant which depends only on the dimension d. Bergstrom [4] found the

optimal value B2 =√5/4, see also §6.5.

Motivated by the Steinitz Lemma, we define a similar notion for integral curves. Let γ = [v1 . . . vn] ∈Mn be an integral curve in R3. We say that γ is B-packing, if |v1vi| ≤ B for all 1 ≤ i ≤ n.Lemma 2.6. Every generic integral curve γ ∈ Gn is flip equivalent to a generic integral curve γ′′ ∈ Gnthat is 3/2-packing.

Here the constant B = 3/2 is chosen somewhat arbitrary. In fact, any constant√5/4 < B <

√3 will

satisfy the lemma and suffice for our purposes.

Proof. Fix ε > 0 and γ ∈ Gn. Let γ′ ∼ γ and ξ = [w1 . . . wn] ∈ Pn be as in the proof of Lemma 2.5, so|γ′, ξ| < ε. Define ui =

−−−−→wiwi+1, 1 ≤ i < n, and un = −−−→wnw1. Clearly, ui are unit vectors which satisfyu1 + . . .+ un = 0. By the Steinitz Lemma, there is a permutation σ ∈ Sn, s.t. (2.6) holds. Consider areduced factorization of σ into adjacent transpositions (i, i+ 1) ∈ Sn:

σ = (kℓ, kℓ + 1) · · · (k1, k1 + 1) ,

where 1 ≤ k1, . . . , kℓ ≤ n− 1, and ℓ = inv(σ) is the number of inversions in σ, see e.g. [32]. This reducedfactorization may not be unique, of course. Denote a partial permutation σj = (kj , kj+1) · · · (k1, k1+1),and the corresponding planar curve ξj = [wσj(1) . . . wσj(n)].

Define a sequence of flips as in (2.2), according to this factorization:

(2.7) γ′ = γ0 →k1γ1 →k2

γ2 →k3. . . →kℓ

γℓ = γ′′ .

Since γj are generic, by Lemma 2.1 and induction, we can always choose the multipliers mj so that|γj , ξj | < ε, 1 ≤ j ≤ ℓ.

Now let ε → 0. As in the proof of Lemma 2.3, by continuity of Φk,m in (2.3), we have the limitplanar curve γ′′ → := [y1 . . . yn] ∈ Pn. By induction on the length ℓ of the factorization, we have:−−−→yiyi+1 = uσ(i), 1 ≤ i < n, and −−→yny1 = uσ(n). By the Steinitz Lemma with Bergstrom constant

B2 =√5/4, we conclude that for sufficiently small ε > 0, the integral curve γ′′ is (B2 + δ)-packing, for

all δ > 0. Taking δ < (3/2−B2), we obtain the result.

Remark 2.7. In notation of the proof above, in a special case of a convex centrally symmetric n-gonξ ∈ Pn, n = 2k, the sequence of unit vectors is u1, . . . , uk,−u1, . . . ,−uk. Take the permutation whichgives the order u1, . . . , uk,−uk, . . . ,−u1; the corresponding limit curve ∈ Pn is then degenerate. Forevery reduced factorization as in the proof, the pattern of rhombi used in the flip sequence then definesa zonotopal tilings, see e.g. [26, Exc. 14.25].

2.8. Proof of Theorem 1.3. We note that if the statement of the theorem is true then it holds with theadditional constraint that v1 = v′1, v2 = v′2, and there is a plane through v′1, v

′2, v

′3 containing v1, v2, v3.

We prove the result by induction. First, closed integral curve of length 3 is a unit triangle, soTheorem 1.3 is trivially true in this case. The case of length 4 is resolved in Lemma 2.2. Note that byLemma 2.4 it suffices to prove the theorem only for generic curves γ = [v1 . . . vn] ∈ Gn.

Let n = 5, and let γ ∈ G5 be a generic integral pentagon. By Lemma 2.6, there is an almost planarcurve γ′ = [w1, . . . , w5] ∈ G5, such that γ′ ∼ γ, and |w1wi| ≤ 3/2 for all 1 ≤ i ≤ 5. If the circumradiusof the triangle w1w3w4 is less than 1, then there is a point z ∈ R3, s.t. |w1z| = |w3z| = |w4z| = 1, seeFigure 4 (left). Otherwise, we make a flip for the vertex w1 using Lemma 2.1 to construct a genericcurve γ′′ = [w′

1, w2, . . . , w5] such that the circumradius of the triangle w′1w3w4 is less than 1 and take a

point z, s.t. |w′1z| = |w3z| = |w4z| = 1. Without loss of generality, we consider the first case only.

Apply now Lemma 2.2 to rhombi ρ1 = [w1w2w3z] and ρ2 = [w1w5w4z], to obtain rhombi ρ′1 =[w1w

′2w3z] ∈ D4 and ρ′2 = [w1w

′5w4z] ∈ D4, which satisfy |w2w

′2|, |w5w

′5| < ε. Attach unit triangle

[w3w4z] to rhombi ρ′1 and ρ′2. This gives the desired pentagon η = [w1w′2w3w4w

′5] ∈ D5, s.t. |γ′, η|F < ε.

Thus, γ′ is reachable. By Lemma 2.3, then so is γ, as desired.

DOMES OVER CURVES 7

We also note that for each ε > 0 there is δ > 0 such that for any angle ψ ∈ [∠w2w1w5−δ,∠w2w1w5+δ],there is γ∗ = [w∗

1w∗2w

∗3w

∗4w

∗5 ] ∈ D5 satisfying |γ′, γ∗|F < ε and ∠w∗

2w∗1w

∗5 = ψ. Indeed, we can just

perturb the construction of η. Fixing w3, w4, and z and multipliers of the rhombi ρ′1 and ρ′2 we canmove w∗

1 in the neighborhood of w1. Then the points w∗2 and w∗

5 defined by w∗1 change continuously

with respect to the position of w∗1 . This means ∠w∗

2w∗1w

∗5 changes continuously as well. It is easy to

check that values both smaller and larger than the initial ∠w2w1w5 are possible so there is an intervalof angles that are possible.

z

w1

γ'

z

w1

w2

w5

w3 w4

η

w6

w7

w2 w5

w3 w4

ρ ρ1 2

γ'

Figure 4. Base of induction n = 5, and step of induction for n = 7.

For n ≥ 6, we employ a similar argument. Let γ ∈ Gn be a generic integral curve as above. ByLemma 2.6, there is an almost planar γ′ = [w1 . . . wn] ∈ Gn, such that γ′ ∼ γ, and |w1wi| ≤ 3/2 for all1 ≤ i ≤ n. We choose a point z ∈ R3 such that |w1z| = |w4z| = 1 and the circumradius of the trianglew2w3z is less than 1. Then for the curve [w1w2w3w4z] we can use the construction from above. Weconclude that in a neighborhood of [w1w2w3w4z] there are curves γ∗ ∈ D5 whose angles w∗

1z∗w∗

4 cover[∠w1zw4 − δ,∠w1zw4 + δ].

Now take ε′ > 0, such that |w1w′1|, |zz′|, |w4w

′4| < ε′ implies ∠w′

1z′w′

4 ∈ [∠w1zw4 − δ,∠w1zw4 + δ].The integral curve γ′ = [w1zw4 . . . wn] has length n − 1 so, by the induction hypothesis, there is γ′′ =[w′

1z′w′

4 . . . w′n] ∈ Dn−1 such that |γ′′, γ′|F < ε′. Since ∠w′

1z′w′

4 ∈ [∠w1zw4 − δ,∠w1zw4 + δ], there is acurve of length five that can be domed in the neighborhood of [w1w2w3w4z] with the same angle. Gluingthe domes of these two curves, we obtain a dome of the curve in the neighborhood of γ. This completesthe proof of the reduction and finishes the proof of the theorem.

3. Regular polygons

3.1. Classical domes. Denote by Qn ⊂ R2 the regular n-gon with unit sides in the xy-plane with thecenter at the origin O. From the introduction, there is a trivial dome over Q3 and Q6, and domes overQ4, Q5 are given by regular pyramids. Less obviously, a tiling of Q12 given in Figure 5 (left), gives anatural dome over Q12, when square pyramids are added.

Figure 5. Left: Tiling giving a dome over Q12. Right: Pentagonal cupola giving a dome over Q10.

Similarly, recall that the regular octagon Q8 and decagon Q10 are spanned by the surfaces of Johnsonsolids square cupola and pentagonal cupola, respectively, see Figure 5 (right) and [17] for details.2 Infact, both are cuts of the Archimedean solids, see e.g. [9, p. 88]. The faces of both surfaces are regulartriangles, squares or pentagons. Adding a pyramid to each face we obtain domes over Q8 and Q10.

2The image on the right is available from the Wikimedia Commons, and is free to use with attribution.

8 ALEXEY GLAZYRIN AND IGOR PAK

3.2. Proof of Theorem 1.4. We follow notation in the proof of Lemma 2.1, and employ the symmetryof Qn at every step.

First, attach a unit triangle to each side of Qn at angle θ > 0 to the plane. In the construction below,we make assumptions on what are distances ai and α defined below. These will prohibit countably manyvalues of θ in a manner similar to the proof of Theorem 1.3. We present the construction first for thepurposes of exposition.

Let the angle θ > 0 be very small, to be chosen at a later point. Denote by a1 the distances betweenvertices of adjacent unit triangles, see Figure 6 below. Assume that a1 /∈ Q. Note that a1 > 0 is welldefined for n ≥ 7.

Moving along the boundaries of triangles attached to Qn, attach to their unit edges n rhombi R1 =ρm1

(a1, ∗

). To simplify the notation, we use (∗) for the second diagonal, since it is completely determined

by the multiplier m1 and a, see the proof of Lemma 2.1. Take m1 large enough and chosen so that R1

is nearly planar, oriented towards the z axis, and at an angle θ1 > θ with the plane. Such m1 exists byLemma 2.1 if we assume further that a1 /∈ Q.

a1

a3

a2

R1

R2

Figure 6. Example of rhombi R1, R2, and distances a1, a2, a3.

Next, moving along the boundary, denote by a2 the distances between vertices of adjacent rhombi R1,and observe that a2 /∈ Q. Now attach to the adjacent unit edges rhombi R2 = ρm2

(a2, ∗

), where the

multiplier m2 is large enough and chosen so that R2 is nearly planar, at an angle θ2 > θ1, see Figure 7.Again, such m2 exists by Lemma 2.1, if we assume further that a2 /∈ Q.

Repeat this procedure for k iterations, until the distance β to the vertical z-axis from new rhombivertices satisfies β <

√1− α2/4. Here α := ak+1 denotes the distance between closest new vertices

of the adjacent rhombi Rk = ρmk

(ak, ∗

), and we assume that that α /∈ Q. The above bound on β

corresponds to having the projection of the nearly planar rhombus Rk+1 cover the origin O, see Figure 7(center).

At this stage, attach to the adjacent unit edges new unit rhombi R = ρM (α, ∗) in such a way thatthe new vertices are at distance δ > 0 from the z-axis, see Figure 7 (center). By Lemma 2.1, distanceδ > 0 can be made as small as necessary.

Now, the construction above is uniquely determined by the angle θ > 0 and the integer sequence

m := (m1,m2, . . . ,mk,M).

Since the number of vectors m is countable, the assumptions ai, α /∈ Q over all m represent countablymany inequalities on θ, so for some θ > 0 the above construction is well defined.

The resulting (partial) surface S is continuously deformed with θ for every fixed m (cf. the proof ofLemma 2.3). Indeed, this follows by induction: first observe that a1 is continuously changing with θ,then so does a2, etc., until we conclude with α; details are straightforward.

Now, continuously changing θ > 0 and using the symmetry, we can place all the remaining free rhombivertices onto the z axis. This completes the construction of a dome over Qn for all n ≥ 7, and the smallercases 3 ≤ n ≤ 6 are discussed above.

Finally, for a regular n-gon rQn, replace unit triangles Q3 with their scaled version rQ3 and proceedas above. Now triangulate every copy rQ3 with r2 unit triangles Q3, completing the construction of adome over rQn.

DOMES OVER CURVES 9

O

Qnθ3

θ1θ2

O

z

δβ

Figure 7. Left: Nearly planar tiling of a portion of Qn with rhombi. Middle: Verticalslice. Right: Example of Q12 with triangles and nearly-flat rhombi R1.

Remark 3.1. Also, one can ask if a version of the arm lemma (see e.g. [26, §22.2]), holds in this case.We believe this to be true for every fixed m, on a sufficiently small interval θ ∈ (θ0 − ǫ, θ0 + ǫ), but thisresult is not necessary for the argument in the proof.

4. The algebra of squared diagonals

4.1. Contractible domes. As a warm-up to the proof of Theorem 1.2, we first present a short argumentfor the case when the spanning surface S is homeomorphic to a disc.

Proposition 4.1. Let γ ⊂ R3 be a unit rhombus γ = ρ(s, t), with diagonal lengths s and t. Suppose γcan be domed by a surface homeomorphic to a disc. Then there exists a polynomial P ∈ Q[x, y], suchthat P (s2, t2) = 0.

For the proof of the proposition, we need to consider doubly periodic surfaces homeomorphic to theplane. Let K be a simplicial connected pure 2-dimensional complex with a free action of the groupG = Z⊕Z with generators a and b. Assume that G acts as a linear bijection on each simplex of K, andthat the number of orbits of triangles under the action of G is finite. Consider a mapping θ : K → R3,linear on each simplex of K, and equivariant with respect to the action of Z ⊕ Z, such that a and bact by translations with vectors α and β, respectively. Then the pair (K, θ) is called a doubly periodictriangular surface. Sometimes, with a slight abuse of notation, we call the surface K as well.

Now, let us construct a doubly periodic surface comprised of unit triangles for every unit triangulationof a unit rhombus.

Lemma 4.2. For a unit rhombus γ = ρ(s, t) with diagonals s and t that can be domed, there is a doublyperiodic surface of unit triangles with two orthogonal periodicity vectors of length s and t, respectively.Moreover, there is such a surface homeomorphic to the plane if the unit triangulation of the rhombusspans a surface homeomorphic to a disc.

Proof. First we construct a doubly periodic surface whose cells are either parallel translates of γ or −γ(see Figure 8). This surface is combinatorially equivalent to a tiling of the plane with unit squares. Achessboard coloring of such a tiling makes white squares correspond to parallel translates of γ and blacksquares correspond to parallel translates of −γ. The periodicity vectors of this surface are the vectors ofthe diagonals of γ.

Attaching a spanning unit triangulation to each translate of γ and −γ we obtain a doubly periodicpolyhedral surface comprised of unit triangles with required periodicity vectors. Clearly, if a spanningtriangulation of γ is homeomorphic to a disk, the resulting doubly periodic surface is homeomorphic tothe plane.

10 ALEXEY GLAZYRIN AND IGOR PAK

γ γ-

Figure 8. Doubly periodic surface from translates of γ and −γ.

At this point we consider only doubly periodic triangular surfaces (K, θ) with unit triangular faces.For a fixed complex K, let G(K) be the set of all possible Gram matrices formed by vectors α and β forall doubly periodic triangular surfaces (K, θ). For a Gram matrix G ∈ G(K), we denote its entries byg11, g12 = g21, and g22.

Theorem 4.3 (Gaifullin–Gaifullin [13]). Let K be a simplicial pure 2-dimensional complex homeomor-phic to R2 with a free action of the group Z ⊕ Z. Then there is a one-dimensional real affine algebraicsubvariety of R3 containing G(K).

In particular, the entries of each Gram matrix G from G(K) satisfy a one-dimensional system of twonon-trivial polynomial equations with integer coefficients:

p(g11, g12, g22) = 0

q(g11, g12, g22) = 0.

Remark 4.4. In fact, the result in [13] is more general, as the authors consider all polygonal doublyperiodic surfaces homeomorphic to the plane with arbitrary sets of side lengths. In this setting, thecoefficients of polynomials p and q are obtained from the ideal generated by squares of all side lengthsof polygons in the polygonal surface

Proof of Proposition 4.1. If a unit rhombus with diagonals s and t can be spanned by unit trianglesthen, by Lemma 4.2, there is a doubly periodic triangular surface with orthogonal periodicity vectors oflength s and t. The entries of the Gram matrix of periodicity vectors are g11 = s2, g12 = 0, g22 = t2.By Theorem 4.3, there are two polynomials p, q with integer coefficients vanishing on the entries of theGram matrix. Thus, at least one of the equations p(s2, 0, t2) = 0 and q(s2, 0, t2) = 0 is non-trivial, andcan be used for the polynomial P .

4.2. Theory of places. There is no result generalizing Theorem 4.3 for doubly periodic surfaces ofnon-trivial topology and, moreover, we will show in Theorem A.2 that such a generalization is not true.However, for our purposes, we do not need two polynomials p and q as in Theorem 4.3. It is sufficientto find at least one polynomial that is non-trivial whenever g12 = 0. The machinery developed in [13]is based on the proof of the bellows conjecture for orientable 2-dimensional surfaces [7], and is also thebasis for our approach. We use places of fields as the main algebraic instrument of the proof.

Let F be a field and F be F extended by ∞, i.e. F = F ∪ ∞ with arithmetic operations extended

to F by

a±∞ =∞ anda

∞ = 0 , for all a ∈ F,

a · ∞ =a

0= ∞ for all a ∈ F \ 0.

The expressions0

0,∞∞ , 0 · ∞ and ∞±∞ are not defined.

Let L be a field. A map φ : L→ F is called a place if φ(1) = 1 and

φ(a± b) = φ(a) ± φ(b), φ(a · b) = φ(a) · φ(b) for all a, b ∈ L,whenever the right-hand side expressions are defined.

DOMES OVER CURVES 11

As a direct consequence of the definition, we have φ(0) = 0 for all places, and φ(x) = ∞ for x 6= 0if and only if φ(−x) = ∞ and if and only if φ(1/x) = 0. It is also clear that whenever charF = 0, wemust also have charL = 0. Similarly, we have φ(kx) =∞ for a non-zero k ∈ Z, if and only if φ(x) =∞.We will use the following basic fact on extensions of places.

Lemma 4.5 (see e.g. [21, Ch. 1, Thm 1]). Let L be a field containing a ring R. Let φ be a homomorphismfrom R to an algebraically closed field Ω, and suppose φ(1) = 1. Then φ can be extended to a placeL→ Ω ∪ ∞.

4.3. General domes. For a doubly periodic triangular surface (K, θ), let α and β be the periodicityvectors of the surface and Λ be the lattice generated by α and β. The number of orbits under the actionof G = Z ⊕ Z is finite, so we choose a representative of each orbit. Let (x1, y1, z1), . . . , (xN , yN , zN ) betheir coordinates in R3, and let (xα, yα, zα), (xβ , yβ , zβ) be the coordinates of the periodicity vectors.Define field L as follows:

L := Q(x1, y1, z1, . . . , xN , yN , zN , xα, yα, zα, xβ , yβ , zβ

).

Note that L does not depend on the choice of representatives of orbits and the choice of the basis for thelattice Λ.

When vertices a, b on the surface (K, θ) form an edge, denote by ℓab the squared distance betweenthem:

ℓab := (xa − xb)2 + (ya − yb)2 + (za − zb)2.Clearly, ℓab ∈ L. For each surface the set of all possible ℓab is finite. Let R be the Q-subalgebra of Lgenerated by all ℓab of the surface.

All vectors in Λ can be written as integer linear combinations of the periodicity vectors, λ = kα+mβ.In case k and m are relatively prime, the vector λ is called primitive. Denote by Λ∗ the set of primitivevectors λ ∈ Λ. By (λ1, λ2) we mean the standard inner product of vectors λ1 and λ2.

As the first step in the proof of Theorem 1.2, we prove the lemma on finite elements of places.

Lemma 4.6 (Main lemma). For a doubly periodic triangular surface (K, θ) obtained by the constructionin Lemma 4.2, let φ : L → F ∪ ∞ be a place that is finite on all ℓab defined by the surface and letcharF = 0. Then there is a vector λ ∈ Λ, λ 6= 0, such that φ is finite on (λ, λ).

For the proof, we use the following technical result of Connelly, Sabitov, and Walz [7, Lemma 4], seealso [26, §34.3].

Theorem 4.7 (Connelly–Sabitov–Walz). Let u be a vertex of a triangular surface in R3 and v1, . . ., vd,d ≥ 4, be adjacent to it in this cyclic order, denote also vd+1 = v1 and vd+2 = v2. Let φ be a place thatis defined on Q(xu, yu, zu, xv1 , yv1 , zv1 , . . . , xvd , yvd , zvd) and is finite on all ℓuvi , ℓvivi+1

, 1 ≤ i ≤ d. Thenφ is finite on at least one of the squared diagonal lengths ℓvivi+2

, 1 ≤ i ≤ d.

4.4. Proof of the Main Lemma 4.6. The statement of the lemma is true if one of the edges ofthe surface forms a vector from Λ. We define the complexity as a partial ordering of doubly periodictriangular surfaces with the same periodicity lattice Λ. Surfaces with edges from Λ are called the leastcomplex (an example is given in Figure 8). For surfaces without edges from Λ, the ordering is defined asfollows.

A surface K1 is said to be less complex than K2, if the Euler characteristic of K1/Λ is greater thanthe Euler characteristic of K2/Λ. The surface K1 is less complex than K2 if the Euler characteristics ofK1/Λ and K2/Λ are the same, and K1/Λ has fewer vertices than K2/Λ. The surface K1 is less complexthan K2 if K1/Λ and K2/Λ have the same Euler characteristic and the same number of vertices, but thesmallest vertex degree of K1 is less than the smallest vertex degree of K2. The proof will proceed byinduction on complexity.

First case. Suppose the surface contains the edges ab, bc, ca, but does not contain a triangle abc. Theclosed curve formed by the edges ab, bc and ca, divides its neighborhood into two components. Then wedefine the surgery along [abc] by removing vertices a, b, c, edges ab, bc, ca, and adding two copies of thetriangle abc, which we call a′b′c′ and a′′b′′c′′. We do this in such a way that a′b′c′ and a′′b′′c′′ retainthe incidences of a, b, c in the first and the second component of the neighborhood, respectively.

12 ALEXEY GLAZYRIN AND IGOR PAK

We make this surgery for all periodic images of abc under the action of Λ. If this surgery keeps thesurface connected, then it increases the Euler characteristic of K/Λ. If the surgery splits the surfaceinto two new surfaces then the Euler characteristic for each of them is not smaller than the initial Eulercharacteristic, for both of them there are fewer vertices than for the initial surface, and at least one ofthem is a connected doubly periodic triangular surface with the periodicity lattice Λ. We call the latterthe connectivity property, see Figure 9.

a'b'

c'

ab

c

Figure 9. Connecticity property in the First Case of a doubly periodic surface.

The connectivity property of one of the two new surfaces requires some explanation. Let T be the torusgenerated by the lattice Λ, i.e. T = R2/Λ. Then, initially, K/Λ ∼= T#D, where # denotes the connectedsum, and D corresponds to the surface formed by the two rhombi γ and −γ and the domes over them.After the transformations to the surface described in the proof of the lemma for the resulting surface K ′,we obtain that K ′/Λ can be always represented as T#D′ for some closed surface D′. The surgerydescribed above preserves T in one of the two disconnected components, thus making its correspondingsurface connected (cf. Remark 4.8 below). Since the set of ℓab for either surface is a subset of the initialset, we can use the inductive step.

Second case. Suppose there are no triples of vertices a, b, c as in the first case. Consider a vertex uof the surface with the smallest degree d adjacent to vertices v1, . . ., vd. The smallest degree must beat least 4, because the first case holds otherwise. We use Theorem 4.7 but we have to be careful withapplying it as the field in Theorem 4.7 is not a subfield of the field L defined earlier. The issue is thatsome vertices vi and vj may belong to the same orbit under the action of the lattice Λ.

Let Ru be a Q-subalgebra of K = Q(xu, yu, zu, xv1 , yv1 , zv1 , . . . , xvd , yvd , zvd) generated by all ℓuvi ,ℓvivi+1

, 1 ≤ i ≤ d. There is a natural homomorphism ψ from Ru to L mapping all elements of Ru totheir corresponding expressions in L. Note that we cannot automatically define this homomorphism onall elements of K. For example, when v3 = v1+α and v6 = v4+α, the image of 1/(xv3 +xv4 −xv1 −xv6)is not defined.

At this point, we use Lemma 4.5 and extend ψ to ψ : K→ L∪∞. The place φ can be also extendedto a place φ : L → F ∪ ∞. In order to construct this extension, we apply Lemma 4.5 to a subring of

all elements of L whose images under φ are finite. For the constructed mapping, φ(x) = 0 if φ(x) = 0.Subsequently, if φ(x) is ∞, φ(x) must be ∞ as well and φ extends the whole place φ.

Using φ(∞) = ∞, we can define the composition φ ψ. This composition is the place from K to

F ∪ ∞. Applying Theorem 4.7 to φ ψ we conclude that there is i such that the composition and,subsequently, φ is finite on ℓvivi+2

.For the next step, we substitute two triangles of the surface, uvivi+1 and uvi+1vi+2 with uvivi+2 and

vivi+1vi+2 simultaneously deleting the edge uvi+1 and adding the edge vivi+2. There was no edge vivi+2

prior to this operation because otherwise the triangle uvivi+2 would satisfy the case considered above. Atthe same time we make the same operations for all triangles that are the images of uvivi+1 and uvivi+2

under the action of Λ. As the result we obtain another surface K ′ such that K ′/Λ is topologically thesame as K/Λ and has the same number of vertices but the minimum vertex degree of K ′ is smaller. Theplace φ is still finite on all ℓab for edges ab, so all conditions of the lemma still hold.

DOMES OVER CURVES 13

Observe that the operations in both cases decrease the complexity of the surface. Note that thiscannot continue indefinitely since the Euler characteristic is at most 2, and the number of edges andvertex degrees are positive. Thus, at some point we reach the least complex surface for which thestatement of the lemma is true.

Remark 4.8. In the proof of the First Case, the connectivity property fails for general doubly periodicsurfaces. In particular, if the elements of the fundamental group of K/Λ corresponding to periodicityvectors do not commute, two new surfaces may be both disconnected unions of one-periodic pieces. Anexample is given in Figure 10. Here we show only the bottom half of the surface, which has connectedcomponents periodic along α. The top half is attached to the bottom along red triangles and has similarstructure, but with connected component periodic along β. This observation will prove crucial in theproof of Theorem A.2 in the Appendix.

β

α

Figure 10. Non-example to the connectivity property in the First Case for generaldoubly periodic surfaces.

4.5. Proof of Theorem 1.2. Let R′ be the Q-subalgebra of L obtained by adding all (λ, λ)−1, λ ∈ Λ∗,to the subalgebra R:

R′ = R[(λ, λ)−1

∣∣ λ ∈ Λ∗ ] .Let I ′ be the following ideal in R′:

I ′ =((λ, λ)−1

∣∣ λ ∈ Λ∗)⊳ R′.

Assume that I ′ 6= R′. Then, by Krull’s theorem (see e.g. [2]), there exists a maximal ideal I, such thatI ′ ⊂ I. Let F = R′/I. Since R′ contains Q, field F must contain Q as well and charF = 0. Let F bean algebraic closure of F . The quotient homomorphism R′ → F satisfies the conditions of Lemma 4.5for Ω = F so it can be extended to the place φ : L → F ∪ ∞. The quotient homomorphism is equalto 0 on all (λ, λ)−1, λ ∈ Λ∗. Therefore, the place φ is infinite on (λ, λ) for all λ ∈ Λ∗. This implies thatthe same holds for all non-zero λ ∈ Λ. On the other hand, the quotient homomorphism is finite on R′.Therefore, we get a contradiction with Lemma 4.6. We conclude that the assumption that I ′ 6= R′ isfalse.

From above, we have that I ′ = R′. In particular, this implies that 1 ∈ I ′:

(4.1) 1 =

M∑

i=1

ri(λi1, λi1)(λi2, λi2) . . . (λipi

, λipi),

where all λij ∈ Λ∗, all ri ∈ R, and pi ≥ 1. After multiplying by the least common multiple of alldenominators, the left hand side of (4.1) becomes

Z :=

N∏

j=1

(λj , λj) =

N∏

j=1

(kjα+mjβ, kjα+mjβ

)=

N∏

j=1

(k2j (α, α) + 2kjmj(α, β) +m2

j(β, β)),

where λj = kjα+mjβ.In the same manner we can write down the products in the right hand side of (4.1) times Z. We rewrite

the equation via the entries of the Gram matrix of the lattice Λ, which are equal to g11 = (α, α) = s2,

14 ALEXEY GLAZYRIN AND IGOR PAK

g22 = (β, β) = t2, and g12 = (α, β) = 0. We also use the fact that polynomial functions ri ∈ R take onlyrational values on doubly periodic unit triangular surfaces, and denote by qi ∈ Q the value of ri on thesurface (K, θ). We then have:

(4.2)

N∏

j=1

(k2j s

2 +m2j t

2)−

M∑

i=1

qi

Ni∏

j=1

(k2ijs

2 +m2ijt

2)= 0 ,

where all Ni < N . Note that this is the only time in the proof we use the fact that we have unit triangles,and that the periodicity vectors are orthogonal.

We conclude that the polynomial P (s2, t2) formed by the equation (4.2) has rational coefficients. Letx← s2 and y ← t2. From above, the polynomial P (x, y) is nonzero since

deg

N∏

j=1

(k2j x + m2

j y)= N,

and the degree of all other terms in (4.2) have degrees Ni < N . This completes the proof.

4.6. Further applications. Note that polynomials P found in the proof of Theorem 1.2 are quitespecial. In some cases, with a more careful analysis, one can conclude non-existence of domes for somerhombi whose diagonals are algebraically dependent over Q. For example, consider the rhombi whoseratio of diagonal lengths is algebraic:

Corollary 4.9. Let s /∈ Q and t/s ∈ Q. Then the unit rhombus ρ(s, t) cannot be domed.

For example, the corollary implies that ρ(1π, 1π

)and ρ

(e√7, e√

8

)cannot be domed.

Proof. Suppose ρ(s, t) ∈ D4, and let c := t/s ∈ Q. Consider a polynomial P (s2, t2), as in the proof ofTheorem 1.2. Viewed as a polynomial in x = s2 over Q the leading degree term of P becomes

N∏

j=1

(k2j x + m2

j c2x

),

so it still has a higher degree than all other terms. Therefore, s ∈ Q, a contradiction.

The following is a generalization of the previous corollary:

Corollary 4.10. Let s /∈ Q, and let s2 and t2 be algebraically dependent with an irreducible polynomialQ(s2, t2) = 0. Suppose Q∗ is the highest degree component of Q. Then the unit rhombus ρ(s, t) cannotbe domed unless Q∗ is (up to a constant) a product of linear polynomials (k2x +m2y), where k,m arenon-negative integers. Moreover, the rhombus can be domed only if any monomial of Q divides one ofthe monomials of Q∗.

For example, the corollary implies that ρ(1π, 1π2

)cannot be domed. Indeed, the irreducible polynomial

x2−y does not satisfy the last condition of Corollary 4.10 since the monomial y does not divide a highestdegree monomial.

Proof. Suppose ρ(s, t) can be domed. We note that (s2, t2) is a root of both P from the proof ofTheorem 1.2 and the irreducible polynomial Q. Either P is divisible by Q, or there are polynomialsA,B ∈ Q[x, y] and a nonzero polynomial D ∈ Q[x] such that AP + BQ = D (see the argument in [12,§1.6, Prop. 2]). The latter case contradicts our assumption on s. We conclude that P is divisible by Q.Therefore, the highest degree component of P given by

N∏

j=1

(k2j x + m2

j y)

must be divisible by Q∗, as desired.Assume now there is a monomial of Q that does not divide any monomial of Q∗. As we already

know, Q∗ is a product of xuyv and linear polynomials (k2x+m2y), where k,m are positive integers. IfdegQ = n, then Q∗ contains all monomials xqyr such that q+r = n and q ≥ u, where r ≥ v. A monomialof Q that does not divide any monomial of Q∗ must be either divisible by xn−v+1 or yn−u+1. Without

DOMES OVER CURVES 15

loss of generality assume the former. Then the largest degree of x in Q is attained on a monomial notfrom Q∗. Since P is divisible by Q, the largest degree of x in P is attained on a monomial not from thehighest total degree component of P . This clearly contradicts the composition of P in (4.2).

Remark 4.11. The approach in the corollaries fails in two notable cases we cover in the next section.First, by Proposition 5.2 below, an isosceles triangle ∆ with side lengths (2, 2, 1) can be domed if and

only if a unit rhombus ρ(12 ,√3) can be domed. Since the argument above is not applicable for ρ(s, t)

for which s2, t2 ∈ Q, we cannot conclude that ∆ cannot be domed, cf. Conjecture 5.1.The second example where the above approach is inapplicable is the case of planar rhombi ρ

(s, t

),

where s2 + t2 = 4, see §5.3. In fact, one of the product terms k2j s2 +m2

j t2 = (k2j −m2

j)s2 + 4m2

j of theleading degree term in P can be equal to 4 when kj = ±1 and mj = ±1.

5. Big picture

5.1. Integer-sided triangles. It may seem from the proof of Theorem 1.2, that only integral curveswith non-algebraic diagonals cannot be domed. In fact, we believe that only very few integral curveswith algebraic diagonals can be domed.

Conjecture 5.1. An isosceles triangle ∆ with side lengths (2, 2, 1) cannot be domed.

As with many other domes on curves problems, this conjecture turned out to be equivalent to thatover a certain unit rhombus (cf. §5.5 below).

Proposition 5.2. Let ∆ be an isosceles triangle with side lengths (2, 2, 1), and let ρ⋄ = ρ(12 ,√3). Then

∆ can be domed if and only if ρ⋄ can be domed.

Proof. Attach three unit triangles to ∆ as in Figure 11. Observe that the boundary of the resultingsurface is exactly ρ⋄.

∆ ρ

Figure 11. Proof of Proposition 5.2: ∆ ∈ D5 if and only if ρ⋄ ∈ D4.

In a contrapositive fashion, let us show that if ∆ ∈ D5, then all integral triangles can be domed.

Proposition 5.3. Let ∆ be an isosceles triangle with side lengths (2, 2, 1). If ∆ can be domed, then socan every integer-sided triangle.

Proof. Whenever clear, we denote polygons with their edge length sequence. Observe that all triangles(k, k, k) and all isosceles trapezoids (1, ℓ, 1, ℓ+ 1) can be domed by a plane triangulation. To constructdomes over all integer-sided triangles, we use the following rules:

(1) for integer k > 1, 1 ≤ ℓ <√3k, two copies of (k, k, 1), one (k, k, ℓ), and a (1, ℓ, 1, ℓ+ 1)

trapezoid, give a triangle (k, k, ℓ+ 1), via a pyramid over a trapezoid (see Figure 12).

(2) for integer k > 1, ℓ <√3k, two copies of (k, k, ℓ), and one (k, k, 1), give a triangle

(ℓ, ℓ, 1) via a tetrahedron.

We now construct all triangles (k, k, 1) one by one, alternating the rules above in the following order:

∆ = (2, 2, 1)→(1) (2, 2, 3)→(2) (3, 3, 1)→(1) (3, 3, 4)→(2) (4, 4, 1)→ . . .

Next, we construct domes over general isosceles triangles (k, k, ℓ), for all 1 ≤ ℓ < k, as follows:

(k, k, 1)→(1) (k, k, 2)→(1) (k, k, 3)→(1) . . .

Finally, we can span (a, b, c) using triangles (k, k, a), (k, k, b) and (k, k, c), for k ≥ maxa, b, c largeenough. This completes the proof.

16 ALEXEY GLAZYRIN AND IGOR PAK

2

Figure 12. Construction in the first rule: (2, 2, 1)→(1) (2, 2, 3).

Remark 5.4. Suppose, contrary to Conjecture 5.1, that a triangle (2, 2, 1) can be domed. That wouldeasily imply Theorem 1.4. Indeed, let ℓ be an integer greater than the radius of rQn. By Proposition 5.3,triangle (ℓ, ℓ, r) can also be domed. Symmetrically attach these triangles to all edges in rQn, to form apyramid over rQn.

5.2. Flexible surfaces. Let S ⊂ R3 be a PL-surface homeomorphic to a sphere, and whose faces areunit triangles. We say that S is a closed dome. Such S is called flexible, if there is a continuous familySt, t ∈ [0, 1] of (intrinsically) isometric but non-congruent closed domes; closed dome S is called rigidotherwise.

We say that an integral curve is degenerate if it has two vertices that coincide. Let γ ⊂ S be aclosed non-degenerate integral curve along the edges of S. We say that γ is a separating belt if S r γ isdisconnected. We say that a dome S flexes γ, if S is flexible with a continuous family γt ⊂ St, t ∈ [0, 1]of non-congruent integral curves.

Conjecture 5.5. Let S be a closed flexible dome, and γ ⊂ S be a non-degenerate separating belt. ThenS does not flex γ.

Curiously, this general conjecture implies Conjecture 5.1, which at first glance might seem unrelated.

Proposition 5.6. Conjecture 5.5 implies Conjecture 5.1.

Proof. By contradiction, suppose Conjecture 5.1 is false. In other words, suppose triangle ∆ with sidelengths (2, 2, 1) can be domed. By Proposition 5.3, then so can every integer-sided triangle, includingtriangles with sides (3, 7, 7) and (4, 7, 7), respectively. Four copies of each triangle can be attached toform a flexible Bricard octahedron (see e.g. [26, §30.4]), refuting Conjecture 5.5.

5.3. Planar unit rhombi. Denote by A the set of all a ≥ 0 for which the planar unit rhombusρ(a,√4− a2) can be domed. It follows from Lemma 2.1 that X ⊆ A, so A is infinite, where X is

defined in (2.1).

Conjecture 5.7. Set A is countable.

The following result is our only evidence in favor of this conjecture.

Proposition 5.8. Conjecture 5.5 implies Conjecture 5.7.

Proof. By contradiction, suppose Conjecture 5.7 is false. Since there is a countable number of combina-torial types of triangulated surfaces with quadrilateral boundary, there is one type with infinitely manyboundary planar rhombi. Suppose coordinates of the vertices of a rhombus are (±s/2, 0) and (0,±t/2).The space of surfaces of this particular combinatorial type whose boundary is a planar rhombus is analgebraic setM for s, t and coordinates of all other vertices of a dome.

As an algebraic set,M must have a finite number of connected components. Since there are infinitelymany values of s for points of M, there are two values s1 and s2 corresponding to two points in thesame component. Then a path in this component between these two points provides a flex of a dome.Attaching two copies of a dome along the rhombus boundary, gives a nontrivial deformation of the closeddome. This refutes Conjecture 5.5.

Finally, by analogy with Conjecture 5.1, we believe the following claim.

DOMES OVER CURVES 17

Conjecture 5.9. We have: 1/2 /∈ A. In other words, the planar unit rhombus ρ♦ := ρ(1/2,√15/2)

cannot be domed.

There is a nice connection between these two conjectures.

Proposition 5.10. If the planar integral rhombus 2ρ♦ cannot be domed, then both Conjecture 5.1 andConjecture 5.9 are true.

Proof. Observe that 2ρ♦ can be tiled with four copies of ρ♦. Similarly, 2ρ♦ can be tiled with two copiesof the triangle ∆ with sides (2, 2, 1). This implies the result.

5.4. Space colorings. Denote by R3 a unit distance graph of R3, i.e. a graph with vertices points in R3

and edges pairs (x, y) ∈ R3 × R3 such that |xy| = 1. Questions about colorings of R3 avoiding certainsubgraphs are the main subject of the Euclidean Ramsey Theory, see e.g. [14, 30].

Denote by χ : R3 → 1, 2, 3 a coloring of R3. We say that [xyz] ⊂ R3 is a rainbow triangle in χ, if[xyz] is a unit triangle, and χ(x) = 1, χ(y) = 2, χ(z) = 3.

Proposition 5.11. Let ρ = [uvwx] ⊂ R3 be a unit rhombus. Suppose there is a coloring χ : R3 →1, 2, 3 with no rainbow triangles, and such that χ(u) = χ(v) = 1, χ(w) = 2, χ(x) = 3. Then ρ cannotbe domed.

Proof. By contradiction, suppose S is a 2-dimensional triangulated surface with the boundary ∂S = ρ.Consider a closed 2-manifold M := S ∪ [uvx] ∪ [vwx]. By Sperner’s Lemma for closed 2-manifolds,see [25, Cor. 3.1], the number of rainbow triangles in M is even. Note that triangle [vwx] is rainbow,while triangle [uvx] is not. Thus S has at least one rainbow triangle, a contradiction.

The idea to use the coloring to prove that some curves cannot be domed can be illustrated with thefollowing conjecture:

Conjecture 5.12. Let ρ♦ = [uvwx] ⊂ R3 be a unit rhombus defined above. Then there is a coloring asin Proposition 5.11.

5.5. Domes over multi-curves. One can generalize Kenyon’s Question 1.1 to a disjoint union ofintegral curves Υ = γ1 ∪ . . . ∪ γk, and ask for a dome over Υ. A special case of this, when Υ is union oftwo triangles is especially important in view of the Steinhaus problem, see §6.5. It would be interestingif the theory of places can be applied to the following problem:

Conjecture 5.13. There are unit triangles ∆1,∆2 ⊂ R3, such that Υ = ∆1 ∪∆2 cannot be domed.

In the spirit of Proposition 5.2 and the proof of Theorem 1.3, we conjecture that for every integralcurve γ ⊂ R3, whether it can be domed can be reduced to a single rhombus. This is the analogue of“cobordism for domes”. Formally, in the notation above, we believe the following holds:

Conjecture 5.14. For every integral curve γ ∈ Mn, there is a unit rhombus ρ ∈ M4, and a dome overγ ∪ ρ.

In the spirit of the proof Theorem 1.3, there is a natural way to split Conjecture 5.14 into two parts.

Conjecture 5.15. For every integral curve γ ∈Mn, there is a finite set of unit rhombi ρ1, . . . , ρk ∈ M4,and a dome over γ ∪ ρ1 ∪ · · · ∪ ρk.

Conjecture 5.15 is of independent interest. If true, it reduces Conjecture 5.14 to the following claim:

Conjecture 5.16. For every two unit rhombi ρ1, ρ2 ∈ M4, there is a unit rhombus ρ3 ∈ M4 and adome over ρ1 ∪ ρ2 ∪ ρ3.

18 ALEXEY GLAZYRIN AND IGOR PAK

5.6. General algebraic dependence. While much of the paper and earlier conjectures are largelyconcerned with reducing the problem to domes over rhombi, there is another direction one can explore.Namely, one can ask if Theorem 1.2 can be generalized to all integral curves.

Let γ = [v1 . . . vn] ∈ Mn be an integral curve. Denote by Ln = Q[x1,3, x1,4, . . . , xn−2,n] the ringof rational polynomials with variables corresponding to diagonals of γ. Let CMn ⊂ Ln be the idealspanned by all Cayley–Menger determinants on vertices v1, . . . , vn, see [7] and [26, §41.6]. We cannow formulate the conjecture.

Conjecture 5.17. Let γ = [v1 . . . vn] ∈ Dn be an integral curve which can be domed, where n ≥ 5.Denote by dij = |vivj | the diagonals of γ, where 1 ≤ i < j ≤ n. Then there is a nonzero polynomialP ∈ Ln, which does not belong to the radical of CMn and such that P

(d21,3, d

21,4, . . . , d

2n−2,n

)= 0.

This conjecture can be viewed as a direct analogue of Sabitov’s theory of volume being algebraicover squared diagonal lengths, see §6.6. It would be interesting to see if this result can be obtained byexpanding our argument in Section 4. Perhaps, Conjecture 5.14 could be used to deduce Conjecture 5.17from Theorem 1.2.

6. Final remarks

6.1. Our choice of terminology “dome over curve γ” owes much to the architectural style of the iconicgeodesic domes popularized by Buckminster Fuller, and his ill-fated 1960 proposal of a Dome overManhattan, see e.g. [5, pp. 321–324].

6.2. Kenyon formulated Question 1.1 in [20, Problem 2], in an undated webpage going back to at leastApril 2005. It is best understood in the context of regular polygonal surfaces (see e.g. [1, 8]). Whilewe study only the weaker notion (realizations), both the immersed and the embedded surfaces can beconsidered, as they add further constraints to the domes. Note also that combinatorially, a dome is aunit distance complex of dimension two [18], a notion generalizing the unit distance graphs in §5.4.6.3. One can view the proof of Theorem 1.2 as a rigidity result for 2-surfaces with unit triangular facesand a single boundary. There are several related rigidity results for surfaces with square and regularpentagonal faces, see e.g. [1, 10].

6.4. It is quite possible that Conjecture 5.12 is false while Conjecture 5.9 is true, since the former seemsmuch stronger. This conjecture is partly motivated by our early attempts to use Monsky’s valuationapproach [23, 24], to obtain a negative answer to Kenyon’s Question 1.1.

6.5. The Steinitz Lemma mentioned in §2.7 is a special case of the remarkable 1913 result by Steinitz,motivated by Riemann’s study of conditionally convergent series of real numbers. Bergstrom’s lowerbound B2 ≥

√5/4 comes from taking unit edge vectors in the (k, k, 1) triangle, while the matching

upper bound is based on elementary arguments in plane geometry. For general d, the best known boundBd ≤ d is due to Grinberg and Sevast’janov [15]. Barany and others conjecture that Bd = O(

√d), which

would match the Bergstrom–type lower bound Bd ≥√(d+ 3)/4. We refer to an interesting survey [3]

for these results and further references.

6.6. Building on his earlier work and on [7], Sabitov in [27] and [28, §14], proved that a small diagonalin a closed orientable simplicial polyhedron (of any genus), depends algebraically on the lengths of edgesof the polyhedron and this dependence is generically non-trivial. Following the proof of Theorem 1.2, wecan extend this result to non-orientable polyhedra.

6.7. While Theorem 1.3 is technical, it is natural in view of the existing recreational literature. Notably,in the Scottish book, Steinhaus introduced the tetrahedral chains, which are polyhedra with a chain-likepartition into regular unit tetrahedra. They can be viewed as special types of domes over two triangles,see §5.5. Steinhaus’s 1957 problem asks if tetrahedral chains can be closed, and if they are dense in R3.While the former was given a negative answer in 1959 by Swierczkowski, the latter was partially resolvedonly recently by Elgersma and Wagon [11]. A somewhat stronger version was later proved by Stewart [33].

Stewart’s paper is especially notable. He uses the ergodic theory of non-amenable group actions, andreproves the (previously known) fact that as a subgroup of O(3,R), the group G of face reflections of aregular tetrahedron is isomorphic to a free product: G ≃ Z2 ∗Z2 ∗Z2 ∗ Z2. From there, Stewart showed

DOMES OVER CURVES 19

that G is dense in O(3,R). One can view this result as an advanced generalization of our Lemma 2.2.The original Steinhaus problem about the group of reflections being dense in the full group O(3,R)⋉R3

of rigid motions remains open.

Acknowledgements. We are grateful to Arseniy Akopyan, Nikolay Dolbilin, Alexander Gaifullin, OlegKarpenkov, Oleg Musin, Bruce Rothschild and Ian Stewart, for interesting comments and help with thereferences. We are especially thankful to Rick Kenyon for encouragement, and to anonymous referee forcareful reading of the paper. The authors were partially supported by the NSF.

References

[1] I. M. Alevy, Regular polygon surfaces, preprint (2018), 25 pp.; arXiv:1804.05452.[2] M. F. Atiyah and I. G. Macdonald, Introduction to commutative algebra, Addison-Wesley, Reading, MA, 1969.[3] I. Barany, On the power of linear dependencies, in Building bridges, Springer, Berlin, 2008, 31–45.[4] V. Bergstrom, Zwei Satze uber ebene Vectorpolygone (in German), Abh. Math. Sem. Univ. Hamburg 8 (1931),

148–152.[5] E. Bookstein, The Smith Tapes, Princeton Arch. Press, New York, 2015, 416 pp.

[6] R. Connelly, A counterexample to the rigidity conjecture for polyhedra, IHES Publ. Math. 47 (1977), 333–338.[7] R. Connelly, I. Sabitov and A. Walz, The bellows conjecture, Beitr. Algebra Geom. 38 (1997), 1–10.[8] H. S. M. Coxeter, Regular polytopes (Second ed.), Macmillan, New York, 1963.[9] P. R. Cromwell, Polyhedra, Cambridge Univ. Press, New York, 1997.

[10] N. P. Dolbilin, M. A. Shtan’ko and M. I. Shtogrin, Nonbendability of the covering of a sphere by squares, Dokl. Akad.

Nauk 354 (1997), no. 4, 443–445.[11] M. Elgersma and S. Wagon, Closing a platonic gap, Math. Intelligencer 37 (2015), no. 1, 54–61.[12] W. Fulton, Algebraic curves. An introduction to algebraic geometry, W. A. Benjamin, New York, 1969, 226 pp.[13] A. A. Gaifullin and S. A. Gaifullin, Deformations of period lattices of flexible polyhedral surfaces, Discrete Comput.

Geom. 51 (2014), 650–665.[14] R. L. Graham, Euclidean Ramsey theory, in Handbook of discrete and computational geometry (Second ed.), CRC

Press, Boca Raton, FL, 2004, 239–254.[15] V. S. Grinberg and S. V. Sevast’janov, Value of the Steinitz constant, Functional Anal. Appl. 14 (1980), 125–126.[16] I. Izmestiev, Classification of flexible Kokotsakis polyhedra with quadrangular base, IMRN 2017 (2017), 715–808.[17] N. W. Johnson, Convex polyhedra with regular faces, Canadian J. Math. 18 (1966), 169–200.[18] G. Kalai, Some old and new problems in combinatorial geometry I: around Borsuk’s problem, in Surveys in combi-

natorics, Cambridge Univ. Press, Cambridge, 2015, 147–174.[19] O. N. Karpenkov, On the flexibility of Kokotsakis meshes, Geom. Dedicata 147 (2010), 15–28.[20] R. Kenyon, Open problems, available at https://gauss.math.yale.edu/˜rwk25/openprobs/index.html; an April 5, 2005

version is cached by the Wayback Machine.[21] S. Lang, Introduction to algebraic geometry, Addison-Wesley, Reading, MA, 1972.[22] J. Li, Flexible Polyhedra: Exploring finite mechanisms of triangulated polyhedra, Ph.D. thesis, University of Cam-

bridge, 2018, 176 pp.; available at https://www.repository.cam.ac.uk/handle/1810/271806 .[23] P. Monsky, On dividing a square into triangles, Amer. Math. Monthly 77 (1970), 161–164.[24] P. Monsky, A conjecture of Stein on plane dissections, Math. Z. 205 (1990), 583–592.[25] O. R. Musin, Extensions of Sperner and Tucker’s lemma for manifolds, J. Combin. Theory, Ser. A 132 (2015),

172–187.[26] I. Pak, Lectures on discrete and polyhedral geometry, monograph draft (2009); available at

http://www.math.ucla.edu/˜pak/book.htm .[27] I. Kh. Sabitov, Algorithmic solution of the problem of the isometric realization of two-dimensional polyhedral metrics,

Izv. Math. 66 (2002), 377–391.[28] I. Kh. Sabitov, Algebraic methods for the solution of polyhedra, Russian Math. Surveys 66 (2011), 445–505.[29] B. Schulze, Combinatorial rigidity of symmetric and periodic frameworks, inHandbook of geometric constraint systems

principles, CRC Press, Boca Raton, FL, 2019, 543–565.[30] A. Soifer, The mathematical coloring book, Springer, New York, 2009.[31] H. Stachel, Flexible polyhedral surfaces with two flat poses, Symmetry 7 (2015), 774–787.[32] R. P. Stanley, Enumerative combinatorics, vol. 1 (second ed.), Cambridge Univ. Press, 2012.[33] I. Stewart, Tetrahedral chains and a curious semigroup, Extracta Math. 34 (2019), 99–122.

20 ALEXEY GLAZYRIN AND IGOR PAK

Appendix A. Doubly periodic surface with a three-dimensional flex

A.1. Flex dimension. In [13], Gaifullin and Gaifullin studied the case of doubly periodic surfaceshomeomorphic to the plane. In this case they proved a stronger result than Theorem 4.3, that there aretwo primitive vectors λ, µ ∈ Λ∗, such that the place φ is finite both on (λ, λ) and (λ, µ). They concludedthe following result:

Theorem A.1 ([13, Thm 1.4]). Every embedded doubly periodic triangular surface homeomorphic to aplane has at most one-dimensional doubly periodic flex.

By a doubly periodic flex of the triangular surface S we mean a continuous rigid deformation St, t ∈[0, δ) for some δ > 0, which preserves double periodicity, i.e. invariant under the action of G = Z ⊕ Z

(the action of G can also depend on t). The continuity of S is meant with respect to all dihedral angles.Here we identify deformations modulo changes of parameter t and ask for the dimension of the space offlexing at t = 0, i.e. when S0 = S. For example, the surface in Figure 9, when triangulated along theshadow lines has only one doubly periodic flex along these lines.

Let us mention that flexible doubly periodic surfaces is an important phenomenon in Rigidity Theory,with Kokotsakis surfaces introduced in 1933, giving classical examples, see e.g. [16, 19]. Note that thereare doubly periodic polyhedral surfaces whose flexes are not doubly periodic, see [31]. We refer to [29,§25.5] for a recent short survey on rigidity of periodic frameworks, and further references.

A.2. New construction. In [13, Question 1.5], the authors asked if Theorem A.1 can be extendedto surfaces which are not homeomorphic to a plane. In this section we give a negative answer to thisquestion by an explicit construction.

Theorem A.2. There is a doubly periodic triangular surface whose doubly periodic flex is three-dimensional.

Proof. First, consider a flexible polyhedron F satisfying the following conditions. Polyhedron F musthave two faces f1 and f2, both of them centrally symmetric, such that the distance between their centerschanges during the flexing and all other faces of F are triangles. To construct such F , take, for example,one of Bricard’s flexible octahedra B (see e.g. [22, §2.3] and [26, §30]), and attach two square pyramidsto either two of its triangular faces. We illustrate this step in Figure 13, where B is replaced with theusual octahedron for clarity.

B F

f1

f2

Figure 13. Illustration of the first step of the construction.

We call the axis of F , the line segment connecting the centers of f1 and f2 (see Figure 13). Let Hbe a flexible polyhedron that has two faces congruent to one of the triangular faces of F . We attach

H

F F21

F

g2

g1

g2

g1

g1

g2

σ

H

Figure 14. Illustration of the second step of the construction.

DOMES OVER CURVES 21

two copies of F , which we call F1 and F2, to each of these two faces of H (see Figure 14). For ourconstruction we are interested in such H that, when flexing H , the angle σ between the axes of F1 andF2 changes and is never zero. Again, a suitable Bricard’s octahedron satisfies this condition. We thenhave a three-dimensional flexing of the whole structure: flexing of F1 changes the length of the axis ofF1, flexing of F2 changes the length of the axis of F2, flexing of H changes the angle σ between the axes.

For the next step of the construction we consider F ′, the image of F under the central symmetry with

respect to the center of f2. By F we denote the union of F and F ′ attached by f2 (see Figure 15). From

now on, we consider only flexes of F such that it stays centrally symmetric with respect to the centerof f2. Note that during the flex, face f1 and its counterpart in F ′, face f ′

1, are translates of each other

and the distance between their centers changes during the flex. One can think of F as a polyhedralversion of accordion with bellows such that the sturdy parts of the accordion always stay parallel butthe distance between them may change.

F'f 2'

f '1

F f2

f1 g

1

g2

H

g1

g2 F

F'

Ff1

f '1

Figure 15. Illustration of the pieces of the infinite accordion.

Using infinitely many copies of F , we construct a periodic flexible surface S (infinite accordion) by

attaching f1 of one copy of F to f ′1 of the next copy of F . See Figure 16 for an illustration (cf. Figure 9).

The space of periodic flexes of S is one-dimensional.

Consider a flex St of S = S0 with periodicity vector α. At one of the flexed copies of F we attach H

along g1. Then attach H along g2 to another copy of F ′ which forms its own copy S′ of the infiniteaccordion. Assume the vector of periodicity of S′ is β. Now we attach all translates H + kα, k ∈ Z, tosurface S, and attach S′ + kα to each of them. Then attach H + kα+mβ, m ∈ Z, to all S′ + kα, andattach all translates S +mβ to all translates of H . The resulting surface F is doubly periodic. It canbe flexed in the following two ways:

by changing lengths of both α and β when flexing St and S′z, respectively, and

by changing the angle σ between the axes of S and S′, by simultaneously flexing all copies of H.

Therefore, the space of doubly periodic flexes of F is three-dimensional.

S

F

F

F'

F'

F

F

F'

F'

S'

S+β

S'+α

Figure 16. Illustration of the infinite accordion surface S in the proof, and the threeother copies S + β, S′ and S′ + α, of the doubly periodic surface F .

22 ALEXEY GLAZYRIN AND IGOR PAK

Remark A.3. Note that the surface S in the proof is not necessarily embedded. One can similarlyconstruct the analogous embedded surface, by a more careful choice of a flexible polyhedron, cf. [6].It would be interesting to see if such constructions can have engineering applications. We refer to arecent thesis [22], which reviews several new constructions of embedded flexible polyhedra with largerflex dimensions, and discusses various applications.

Remark A.4. This surface F is a counterexample to a natural generalization of the Main Lemma 4.6.Let us mention why the proof of the Main Lemma fails for F . Note that when the initial polyhedraF and H are homeomorphic to a sphere, we have K/Λ is a surface of genus 2 (in the notation of theproof of the Main Lemma), where the elements of the fundamental group corresponding to α and β donot commute, i.e. stand for two different handles of the surface. In particular, the inductive step in theFirst Case of the proof of the Main Lemma would not work for surgeries since cutting along t woulddisconnect all translates of S1 and all translates of S2, see Remark 4.8.